As metals are giant lattice structures, the number of electrostatic forces to be broken is extremely large, and so metals have high melting and boiling points. This means that the melting point and boiling point of metals are more similar to those for ionic compounds than for covalent substances.

Explanation:

3 0

3 years ago

Filtration and dialysis are techniques that rely on differences in particle size to separate mixture components. True False

soldier1979 [14.2K]

This is true. Think about when you make pasta. The colander acts as a filter and doesn't let the past a through but allows the extra water to escape because the pasta is too big to fit.

3 0

3 years ago

Read 2 more answers

How to separate sand, pebbles, toothpicks and paper clips?<br>plzzzzzzzzzzzzzzzzzzzzzzzz

Step2247 [10]

First, by magnet separate paper clips.
Then, By handpicking, separate the pebbles.
Then, using a suitable sieve, separate toothpicks and toothpicks.

6 0

3 years ago

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

4 years ago